Sequel to Group Theory

We'll start with reminding the reader about the

In the previous document we considered the cyclic group C_{12} in its realization of the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} under addition modulo 12. The initial version of that realization was such that the elements were arranged in there natural order. See the previous document Table 5.1. Then we rearranged the elements such that the subgroup H = {0, 6} stood out. We discovered that this subgroup supported a **homomorphism** **:** The set of **cosets** of this subgroup (including the subgroup itself) was itself a group. This group then was accordingly the image group of the homomorphism associated with this subgroup. Moreover this subgroup was mapped by the homomorphism onto the identity element of the image group. It was called the **kernel** of that homomorphism.

We now consider yet another rearrangement of the elements of the group, such that now the subgroup **P** = {0, 4, 8} will stand out. Indeed, when we try to generate the *whole* group with the element 4, namely by a repeated application of it we find **:**

4 + 4 (mod 12) = 8, i.e. now element 8 is generated.

4 + 4 + 4 (mod 12) = 0, i.e. now the element 0 is generated.

4 + 4 + 4 + 4 (mod 12) = 4, i.e. we are back at element 4.

When we try to generate the *whole* group with the element 8, we get **:**

8 + 8 (mod 12) = 4.

8 + 8 + 8 (mod 12) = 0.

8 + 8 + 8 + 8 (mod 12) = 8.

So we can only generate the elements 0, 4, and 8. We stay within the subset {0, 4, 8}, and thus this subset is *closed* under the group operation *addition modulo 12*. It contains the *identity element*. And because

0 + **0** (mod 12) = 0 (where 0 is element of the subset),

4 + **8** (mod 12) = 0 (where 4, 8 and 0 are elements of the subset), and

8 + **4** (mod 12) = 0 (where 8, 4 and 0 are elements of the subset),

every element of the subset {0, 4, 8} has an *inverse*.

*Associativity* is guaranteed because we still have to do with addition modulo 12 which is associative. So now we are sure that the subset **P** = {0, 4, 8} is a sub*group*.

The group table of our group C_{12} in its realization in finite arithmetic, and with its elements rearranged such that the subgroup (0, 4, 8} and its cosets stands out, is as follows **:**

+ mod 12 |
0 |
4 |
8 |
1 |
5 |
9 |
2 |
6 |
10 |
3 |
7 |
11 |
period |

0 |
0 |
4 |
8 |
1 |
5 |
9 |
2 |
6 |
10 |
3 |
7 |
11 |
1 |

4 |
4 |
8 |
0 |
5 |
9 |
1 |
6 |
10 |
2 |
7 |
11 |
3 |
3 |

8 |
8 |
0 |
4 |
9 |
1 |
5 |
10 |
2 |
6 |
11 |
3 |
7 |
3 |

1 |
1 |
5 |
9 |
2 |
6 |
10 |
3 |
7 |
11 |
4 |
8 |
0 |
12 |

5 |
5 |
9 |
1 |
6 |
10 |
2 |
7 |
11 |
3 |
8 |
0 |
4 |
12 |

9 |
9 |
1 |
5 |
10 |
2 |
6 |
11 |
3 |
7 |
0 |
4 |
8 |
4 |

2 |
2 |
6 |
10 |
3 |
7 |
11 |
4 |
8 |
0 |
5 |
9 |
1 |
6 |

6 |
6 |
10 |
2 |
7 |
11 |
3 |
8 |
0 |
4 |
9 |
1 |
5 |
2 |

10 |
10 |
2 |
6 |
11 |
3 |
7 |
0 |
4 |
8 |
1 |
5 |
9 |
6 |

3 |
3 |
7 |
11 |
4 |
8 |
0 |
5 |
9 |
1 |
6 |
10 |
2 |
4 |

7 |
7 |
11 |
3 |
8 |
0 |
4 |
9 |
1 |
5 |
10 |
2 |
6 |
12 |

11 |
11 |
3 |
7 |
0 |
4 |
8 |
1 |
5 |
9 |
2 |
6 |
10 |
12 |

Table 6.1

P = {0, 4, 8} is a subgroup. We will determine its

Its left coset by the element 0 is 0P = 0{0, 4, 8} = {0 + 0 (mod 12), 0 + 4 (mod 12), 0 + 8 (mod 12)} = {0, 4, 8}.

Its right coset by the element 0 is P0 = {0 + 0 (mod 12), 4 + 0 (mod 12), 8 + 0 (mod 12)} = {0, 4, 8}. So we can say that

In the same way we can determine all cosets of this subgroup P, and because the operation *addition modulo 12* is commutative, we know that all left and right cosets of P with respect to a particular element of the group are equal, which means that the subgroup P = {0, 4, 8} is a *normal* subgroup in G (which is the whole group, namely our group C_{12}, realized in modulo 12 arithmetic) **:**

4P = 4{0, 4, 8} = {4 + 0 (mod 12), 4 + 4 (mod 12), 4 + 8 (mod 12)} = {4, 8, 0} = P (because, of course, {4, 8, 0} = {0, 4, 8}). So

**4P = P4 = P = {0, 4, 8}**.

8P = 8{0, 4, 8) = {8 + 0 (mod 12), 8 + 4 (mod 12), 8 + 8 (mod 12)} = {8, 0, 4} = P. So

**8P = P8 = P = {0, 4, 8}**.

So the cosets of the subgroup P by its own elements are all equal to P itself.

1P = 1{0, 4, 8} = {1 + 0 (mod 12), 1 + 4 (mod 12), 1 + 8 (mod 12)} = {1, 5, 9}. So

**1P = P1 = {1, 5, 9}**.

5P = 5{0, 4, 8} = {5 + 0 (mod 12), 5 + 4 (mod 12), 5 + 8 (mod 12)} = {5, 9, 1} = 1P. So

**5P = P5 = 1P = P1 = {1, 5, 9}**.

9P = 9{0, 4, 8} = {9 + 0 (mod 12), 9 + 4 (mod 12), 9 + 8 (mod 12)} = {9, 1, 5} = 5P. So

**9P = P9 = 5P = P5 = P1 = 1P = {1, 5, 9}**.

2P = P2 = 2{0, 4, 8} = {2 + 0 (mod 12), 2 + 4 (mod 12), 2 + 8 (mod 12)} = {2, 6, 10}. So

**2P = P2 = {2, 6, 10}**.

6P = P6 = 6{0, 4, 8} = {6 + 0 (mod 12), 6 + 4 (mod 12), 6 + 8 (mod 12)} = {6, 10, 2} = P2. So

**6P = P6 = P2 = 2P = {2, 6, 10}**.

10P = 10{0, 4, 8} = {10 + 0 (mod 12), 10 + 4 (mod 12), 10 + 8 (mod 12)} = {10, 2, 6} = 6P. So

**10P = P10 = 6P = P6 = P2 = 2P = {2, 6, 10}**.

3P = 3{0, 4, 8} = {3 + 0 mod 12), 3 + 4 (mod 12), 3 + 8 (mod 12)} = {3, 7, 11}. So

**3P = P3 = {3, 7, 11}**.

7P = 7{0, 4, 8} = {7 + 0 (mod 12), 7 + 4 (mod 12), 7 + 8 (mod 12)} = {7, 11, 3} = 3P. So

**7P = P7 = 3P = P3 = {3, 7, 11}**.

11P = 11{0, 4, 8} = {11 + 0 (mod 12), 11 + 4 (mod 12), 11 + 8 (mod 12)} = {11, 3, 7} = 7P. So

**11P = P11 = 7P = P7 = 3P = P3 = {3, 7, 11}**

The cosets {0, 4, 8}, {1, 5, 9}, {2, 6, 10} and {3, 7, 11} can be written as **:**

{0, 0 + 4, 0 + 4 + 4},

{1, 1 + 4, 1 + 4 + 4},

{2, 2 + 4, 2 + 4 + 4}, and

{3, 3 + 4, 3 + 4 + 4}.

So we can represent these sets by the numbers 0, 1, 2, and 3, which now themselves form a set C = {0, 1, 2, 3}.

This means that we can set up a second **homomorphism** -- The first homomorphism was set up in the previous document with respect to the subset {0, 6} -- which is a homomorphism with respect to the subgroup {0, 4, 8}, which maps

the subset {0, 4, 8} of our group G onto the element **0** of C,

the subset {1, 5, 9} of our group G onto the element **1** of C,

the subset {2, 6, 10} of our group G onto the element **2** of C, and

the subset {3, 7, 11} of our group G onto the element **3** of C.

If we now replace the subsets (cosets) {0, 4, 8}, {1, 5, 9}, {2, 6, 10} and {3, 7, 11} in the above group table (Table 6.1) by those numbers 0, 1, 2 and 3 , we see that the structure which emerges is that of the arithmetic of addition modulo 4, i.e. a realization of the cyclic group of order 4, **C _{4}**

+ mod 4 |
0 |
1 |
2 |
3 |
period |

0 |
0 |
1 |
2 |
3 |
1 |

1 |
1 |
2 |
3 |
0 |
4 |

2 |
2 |
3 |
0 |
1 |
2 |

3 |
3 |
0 |
1 |
2 |
4 |

Table 6.2

So the group C

We can see that the

+ mod 12 |
0 |
4 |
8 |
1 |
5 |
9 |
2 |
6 |
10 |
3 |
7 |
11 |

0 |
0 |
4 |
8 |
1 |
5 |
9 |
2 |
6 |
10 |
3 |
7 |
11 |

4 |
4 |
8 |
0 |
5 |
9 |
1 |
6 |
10 |
2 |
7 |
11 |
3 |

8 |
8 |
0 |
4 |
9 |
1 |
5 |
10 |
2 |
6 |
11 |
3 |
7 |

1 |
1 |
5 |
9 |
2 |
6 |
10 |
3 |
7 |
11 |
4 |
8 |
0 |

5 |
5 |
9 |
1 |
6 |
10 |
2 |
7 |
11 |
3 |
8 |
0 |
4 |

9 |
9 |
1 |
5 |
10 |
2 |
6 |
11 |
3 |
7 |
0 |
4 |
8 |

2 |
2 |
6 |
10 |
3 |
7 |
11 |
4 |
8 |
0 |
5 |
9 |
1 |

6 |
6 |
10 |
2 |
7 |
11 |
3 |
8 |
0 |
4 |
9 |
1 |
5 |

10 |
10 |
2 |
6 |
11 |
3 |
7 |
0 |
4 |
8 |
1 |
5 |
9 |

3 |
3 |
7 |
11 |
4 |
8 |
0 |
5 |
9 |
1 |
6 |
10 |
2 |

7 |
7 |
11 |
3 |
8 |
0 |
4 |
9 |
1 |
5 |
10 |
2 |
6 |

11 |
11 |
3 |
7 |
0 |
4 |
8 |
1 |
5 |
9 |
2 |
6 |
10 |

+ mod 4 |
0 |
1 |
2 |
3 |

0 |
0 |
1 |
2 |
3 |

1 |
1 |
2 |
3 |
0 |

2 |
2 |
3 |
0 |
1 |

3 |
3 |
0 |
1 |
2 |

The group

We'll begin with the former, namey {0, 3, 6, 9} which we will call

+ mod 12 |
0 |
3 |
6 |
9 |
1 |
4 |
7 |
10 |
2 |
5 |
8 |
11 |
period |

0 |
0 |
3 |
6 |
9 |
1 |
4 |
7 |
10 |
2 |
5 |
8 |
11 |
1 |

3 |
3 |
6 |
9 |
0 |
4 |
7 |
10 |
1 |
5 |
8 |
11 |
2 |
4 |

6 |
6 |
9 |
0 |
3 |
7 |
10 |
1 |
4 |
8 |
11 |
2 |
5 |
2 |

9 |
9 |
0 |
3 |
6 |
10 |
1 |
4 |
7 |
11 |
2 |
5 |
8 |
4 |

1 |
1 |
4 |
7 |
10 |
2 |
5 |
8 |
11 |
3 |
6 |
9 |
0 |
12 |

4 |
4 |
7 |
10 |
1 |
5 |
8 |
11 |
2 |
6 |
9 |
0 |
3 |
3 |

7 |
7 |
10 |
1 |
4 |
8 |
11 |
2 |
5 |
9 |
0 |
3 |
6 |
12 |

10 |
10 |
1 |
4 |
7 |
11 |
2 |
5 |
8 |
0 |
3 |
6 |
9 |
6 |

2 |
2 |
5 |
8 |
11 |
3 |
6 |
9 |
0 |
4 |
7 |
10 |
1 |
6 |

5 |
5 |
8 |
11 |
2 |
6 |
9 |
0 |
3 |
7 |
10 |
1 |
4 |
12 |

8 |
8 |
11 |
2 |
5 |
9 |
0 |
3 |
6 |
10 |
1 |
4 |
7 |
3 |

11 |
11 |
2 |
5 |
8 |
0 |
3 |
6 |
9 |
1 |
4 |
7 |
10 |
12 |

Table 6.3

The cosets of the subset R = {0, 3, 6, 9} are the sets

{0, 3, 6, 9}, {1, 4, 7, 10}, and {2, 5, 8, 11}. Like we did earlier it can be seen from the group table (version d, as depicted right above) that the left and right cosets of R are equal for every paricular element, i.e. xR = Rx, in which x is any element of the group (i.e. our group C

These cosets can be renamed in the following way **:**

{0, 3, 6, 9} -- {0, 0 + 3, 0 + 3 + 3, 0 + 3 + 3 + 3} -- **0**.

{1, 4, 7, 10} -- {1, 1 + 3, 1 + 3 + 3, 1 + 3 + 3 + 3} -- **1**.

(2, 5, 8, 11} -- {2, 2 + 3, 2 + 3 + 3, 2 + 3 + 3 + 3} -- **2**.

These numbers, representing the cosets of R form a set D = {0, 1, 2}.

So now we can set a homomorphism of our group G, with respect to the subgroup R = {0, 3, 6, 9}, which maps

the subset (coset) {0, 3, 6, 9} of our group G onto the element **0** of the set D,

the subset (coset) {1, 4, 7, 10} of our group G onto the element **1** of the set D, and

the subset (coset) {2, 5, 8, 11} of our group G onto the element **2** of the set D.

Let's call this homomorphism **m**.

If we now replace the subsets (cosets) {0, 3, 6, 9}, {1, 4, 7, 10} and {2, 5, 8, 11} in the above group table (Table 6.3) by those numbers 0, 1 and 2, we see that the structure which emerges is that of the arithmetic of addition modulo 3, i.e. a realization of the cyclic group of order 3, **C _{3}**

+ mod 3 |
0 |
1 |
2 |
period |

0 |
0 |
1 |
2 |
1 |

1 |
1 |
2 |
0 |
3 |

2 |
2 |
0 |
1 |
3 |

Table 6.4

Indeed we see that the global structure of C

In the following we summarize this homomorphism, which we call

+ mod 12 |
0 |
3 |
6 |
9 |
1 |
4 |
7 |
10 |
2 |
5 |
8 |
11 |

0 |
0 |
3 |
6 |
9 |
1 |
4 |
7 |
10 |
2 |
5 |
8 |
11 |

3 |
3 |
6 |
9 |
0 |
4 |
7 |
10 |
1 |
5 |
8 |
11 |
2 |

6 |
6 |
9 |
0 |
3 |
7 |
10 |
1 |
4 |
8 |
11 |
2 |
5 |

9 |
9 |
0 |
3 |
6 |
10 |
1 |
4 |
7 |
11 |
2 |
5 |
8 |

1 |
1 |
4 |
7 |
10 |
2 |
5 |
8 |
11 |
3 |
6 |
9 |
0 |

4 |
4 |
7 |
10 |
1 |
5 |
8 |
11 |
2 |
6 |
9 |
0 |
3 |

7 |
7 |
10 |
1 |
4 |
8 |
11 |
2 |
5 |
9 |
0 |
3 |
6 |

10 |
10 |
1 |
4 |
7 |
11 |
2 |
5 |
8 |
0 |
3 |
6 |
9 |

2 |
2 |
5 |
8 |
11 |
3 |
6 |
9 |
0 |
4 |
7 |
10 |
1 |

5 |
5 |
8 |
11 |
2 |
6 |
9 |
0 |
3 |
7 |
10 |
1 |
4 |

8 |
8 |
11 |
2 |
5 |
9 |
0 |
3 |
6 |
10 |
1 |
4 |
7 |

11 |
11 |
2 |
5 |
8 |
0 |
3 |
6 |
9 |
1 |
4 |
7 |
10 |

+ mod 3 |
0 |
1 |
2 |

0 |
0 |
1 |
2 |

1 |
1 |
2 |
0 |

2 |
2 |
0 |
1 |

Finally, we will discuss that homomorphism of C

The next group table (of C

+ mod 12 |
0 |
2 |
4 |
6 |
8 |
10 |
1 |
3 |
5 |
7 |
9 |
11 |
period |

0 |
0 |
2 |
4 |
6 |
8 |
10 |
1 |
3 |
5 |
7 |
9 |
11 |
1 |

2 |
2 |
4 |
6 |
8 |
10 |
0 |
3 |
5 |
7 |
9 |
11 |
1 |
6 |

4 |
4 |
6 |
8 |
10 |
0 |
2 |
5 |
7 |
9 |
11 |
1 |
3 |
3 |

6 |
6 |
8 |
10 |
0 |
2 |
4 |
7 |
9 |
11 |
1 |
3 |
5 |
2 |

8 |
8 |
10 |
0 |
2 |
4 |
6 |
9 |
11 |
1 |
3 |
5 |
7 |
3 |

10 |
10 |
0 |
2 |
4 |
6 |
8 |
11 |
1 |
3 |
5 |
7 |
9 |
6 |

1 |
1 |
3 |
5 |
7 |
9 |
11 |
2 |
4 |
6 |
8 |
10 |
0 |
12 |

3 |
3 |
5 |
7 |
9 |
11 |
1 |
4 |
6 |
8 |
10 |
0 |
2 |
4 |

5 |
5 |
7 |
9 |
11 |
1 |
3 |
6 |
8 |
10 |
0 |
2 |
4 |
12 |

7 |
7 |
9 |
11 |
1 |
3 |
5 |
8 |
10 |
0 |
2 |
4 |
6 |
12 |

9 |
9 |
11 |
1 |
3 |
5 |
7 |
10 |
0 |
2 |
4 |
6 |
8 |
4 |

11 |
11 |
1 |
3 |
5 |
7 |
9 |
0 |
2 |
4 |
6 |
8 |
10 |
12 |

Table 6.5

In the above table we see that the two cosets of the subgroup S = {0, 2, 4, 6, 8, 10} are the sets {0, 2, 4, 6, 8, 10} and {1, 3, 5, 7, 9, 11}. for every element of the group the left and right cosets are equal, i.e. for any element x of our group G the following relation holds

When we look to these two cosets we see that the one consists of EVEN numbers only, while the other consists of ODD numbers only. So we can represent them by this feature, i.e. we can set up the following **homomorphism** of G, with respect to the subgroup S **:**

{0, 2, 4, 6, 8, 10} -- **EVEN**

{1, 3, 5, 7, 9, 11} -- **ODD**

Let's call this homomorphism **q**.

If we now in the above table replace the subset {0, 2, 4, 6, 8, 10} by **EVEN**, and the subset {1, 3, 5, 7, 11} by **ODD**, then we see that the set {EVEN, ODD} forms a group with the structure of C_{2}, i.e. the cyclic group of order 2 **:**

+ |
EVEN |
ODD |
period |

EVEN |
EVEN |
ODD |
1 |

ODD |
ODD |
EVEN |
2 |

Table 6.6

In this group the group operation is addition (+)

(an) EVEN (number) + (an) EVEN (number) is equal to (an) EVEN (number).

(an) ODD (number) + (an) ODD (number) is equal to (an) EVEN (number).

(an) ODD (number) + (an) EVEN (number) is equal to (an) ODD (number).

It is clear that this operation is commutative.

The first and third additions just shown indicate that EVEN is the identity element.

This group is of course **isomorphic** (i.e. fully equivalent) to the group (0, 1}, + mod 2 **:**

+ mod 2 |
0 |
1 |
period |

0 |
0 |
1 |
1 |

1 |
1 |
0 |
2 |

Table 6.7

For easy comparison we summarize the homomorphism

+ mod 12 |
0 |
2 |
4 |
6 |
8 |
10 |
1 |
3 |
5 |
7 |
9 |
11 |

0 |
0 |
2 |
4 |
6 |
8 |
10 |
1 |
3 |
5 |
7 |
9 |
11 |

2 |
2 |
4 |
6 |
8 |
10 |
0 |
3 |
5 |
7 |
9 |
11 |
1 |

4 |
4 |
6 |
8 |
10 |
0 |
2 |
5 |
7 |
9 |
11 |
1 |
3 |

6 |
6 |
8 |
10 |
0 |
2 |
4 |
7 |
9 |
11 |
1 |
3 |
5 |

8 |
8 |
10 |
0 |
2 |
4 |
6 |
9 |
11 |
1 |
3 |
5 |
7 |

10 |
10 |
0 |
2 |
4 |
6 |
8 |
11 |
1 |
3 |
5 |
7 |
9 |

1 |
1 |
3 |
5 |
7 |
9 |
11 |
2 |
4 |
6 |
8 |
10 |
0 |

3 |
3 |
5 |
7 |
9 |
11 |
1 |
4 |
6 |
8 |
10 |
0 |
2 |

5 |
5 |
7 |
9 |
11 |
1 |
3 |
6 |
8 |
10 |
0 |
2 |
4 |

7 |
7 |
9 |
11 |
1 |
3 |
5 |
8 |
10 |
0 |
2 |
4 |
6 |

9 |
9 |
11 |
1 |
3 |
5 |
7 |
10 |
0 |
2 |
4 |
6 |
8 |

11 |
11 |
1 |
3 |
5 |
7 |
9 |
0 |
2 |
4 |
6 |
8 |
10 |

+ mod 2 |
0 |
1 |

0 |
0 |
1 |

1 |
1 |
0 |

This

In the next document we will continue our study of

To continue click HERE

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